`t_(1//4)` can be taken as the time taken for concentration of reactant to drop to `.^(3)//_(4)` of its initial value. If the rate constant for a first order reaction is `K`, then `t_(1//4)` can be written as:

Options

A. `0.10//K`
B. `0.29//K`
C. `0.69//K`
D. `0.75//K`

Explanation

To solve the problem, we need to determine the expression for \( t_{1/4} \) (the time taken for the concentration of a reactant to drop to \( \frac{3}{4} \) of its initial value) for a first-order reaction with a rate constant \( k \).

### Step-by-Step Solution:

1. **Understanding the Concept**:
For a first-order reaction, the rate of reaction is proportional to the concentration of the reactant. The integrated rate law for a first-order reaction is given by:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
where \( [A]_0 \) is the initial concentration and \( [A] \) is the concentration at time \( t \).

2. **Setting Up the Equation**:
We want to find \( t_{1/4} \), which is the time taken for the concentration to drop to \( \frac{3}{4} \) of its initial value. This means that \( [A] = \frac{1}{4} [A]_0 \). Therefore, we can express this as:
\[ [A] = [A]_0 - \frac{3}{4}[A]_0 = \frac{1}{4}[A]_0 \]

3. **Substituting into the Integrated Rate Law**:
Substitute \( [A] = \frac{1}{4}[A]_0 \) into the integrated rate law:
\[ \ln \left( \frac{[A]_0}{\frac{1}{4}[A]_0} \right) = kt_{1/4} \]
Simplifying this gives:
\[ \ln(4) = kt_{1/4} \]

4. **Solving for \( t_{1/4} \)**:
Rearranging the equation to solve for \( t_{1/4} \):
\[ t_{1/4} = \frac{\ln(4)}{k} \]

5. **Using Logarithmic Properties**:
We know that \( \ln(4) = 2 \ln(2) \). Thus, we can express \( t_{1/4} \) as:
\[ t_{1/4} = \frac{2 \ln(2)}{k} \]

6. **Converting to Base 10 Logarithm**:
To express this in terms of base 10 logarithm, we can use the conversion:
\[ \ln(x) = 2.303 \log_{10}(x) \]
Therefore:
\[ t_{1/4} = \frac{2 \cdot 2.303 \log_{10}(2)}{k} \]
This simplifies to:
\[ t_{1/4} = \frac{0.693}{k} \]

### Final Expression:
Thus, the final expression for \( t_{1/4} \) in terms of the rate constant \( k \) is:
\[ t_{1/4} = \frac{0.693}{k} \]